Question

Consider $(1+x+x^2{)}^{2n}=\sum _{r=0}^{4n}{a}_r{x}^r$, where $a_0,a_1,a_2......a_{4n}$ are real numbers and $n$ is a positive integer
The correct statement is

• A. $a_r=a_{n-r},0\le r\le n$
• B. $a_{n-r}=a_{n+r},0\le r\le n$
• C. $a_r=a_{2n-r},0\le r\le 2n$
• D. $a_r=a_{4n-r},0\le r\le 4n$

Answer 1

The correct option is D $a_r=a_{4n-r},0\le r\le 4n$

$(1+x+x^2{)}^{2n}$
$=a_0+a_1{x}^1+a_2{x}^2...+a_{2n-1}{x}^{2n-1}+a_{2n}{x}^{2n}...+a_{4n}{x}^{4n}$
Hence the last term will be $a_{4n}{x}^{4n}$.
Now by property of binomial coefficient we get ${}^n{C}_r={}^n{C}_{n-r}$
Similarly $a_r=a_{N-r}$
Substituting $N=4n$ in the above case we get
$a_r=a_{4n-r}$ where $r\in [0,4n]$